# successor and feasible successor conditions

Mar 16th, 2009

Hi all, can anyone explain to me very simply how this works in my icnd2 book I cannot understand what it is saying very well, if say router B RP is less than routers A's current FD does this make it a feasible successor? I dont understand how this works and what needs to be what to be or not to be a feasible successor, please help me on this?

cheers

Carl

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## Replies

Giuseppe Larosa Mon, 03/16/2009 - 15:12
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Hello Carl,

in simple words:

RB can be a feasible successor for a route X, if RB advertises a metric (reported distance) that is less then current best metric to reach route X from local node.

the best metric is calculated from the point of view of the local node and it is called FD , the reported distance is the metric as calculated from the point of view of RB the candidate feasible successor.

Or even more simple RB can be feasible successor for route X if it is "nearest" to network X then the local node

if so RB can be a backup path to subnet X with no risk of routing loops

Hope to help

Giuseppe

carl_townshend Tue, 03/17/2009 - 04:24

hi there

if RB advertises a RP that is less then RA's FD to a destination, then how come this does not take over as the successor to the route? what condition does it need to take over as a successor, can you explain with a simple scenario, say 2 routers on a lan with 2 wan links to a remote network with also 2 wan links.

Giuseppe Larosa Tue, 03/17/2009 - 06:27
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Hello Carl,

the key point is that the comparison is made between a metric calculated in the local router (FD = feasible distance) and the metric as calculated on RB.

RA--- E1 link --- RC------ 10.10.10/24

|------ 768 kbps link -- RB-

both RC and RB advertise prefix 10.10.10.0/24

RC becomes the successor provides the better metric.

RB can be a feasible successor because it is nearer to 10.10.10.0/24 then the local router RA.

the parameters associated to RA-RB links delay and bandwidth are not used by RB in calculating its distance from 10.10.10.0/24

the feasibility condition just states this: to pass the test the advertised distance has to be lower (less not less equal) then the current best metric to the prefix

the EIGRP metric is cumulative in delay and inverse proportional to lowest bandwidth in path to the network.

so if both RB and RC see 10.10.10.0/24 with metric 1000.

the best metric for RA is that via RC.

whatever link is between RA and RC the metric as seen by RA is greater then 1000.

Hope to help

Giuseppe

carl_townshend Tue, 03/17/2009 - 10:01

Hi

I am still not quite clear

if the AD distance has to be lower than the current FD, then why does it not use this lower distance as the primary route?

can you explain a little more?

cheers

Giuseppe Larosa Tue, 03/17/2009 - 10:16
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Hello Carl,

let's use the example of my second post

net 10.10.10.0/24 is out an ethernet interface

both RB and RC are connected to this subnet with their lan interfaces

bw= 10000 kbps delay= 2000 microsec

this provides an advertised distance of:

256 * 200 + 256 * (10000000/10000) =

256 *200 + 256*1000 = 307200

when RA calculates the metric it needs to take in account the following

the path delay becomes: 2000 + 10000

(let's suppose E1 has delay 10000 and BW 2048)

the lowest BW in path becomes 2048

a new calculation occurs

FD = 256 * (12000) + 256 * (10000000/2048)

the path via RB cannot be successor because in this case the bandwith component is only 768 kbps

Distance_via_RB =

256 * (12000) + 256 * (10000000/768)

that is higher.

but AD is lower then FD so RB can be a feasible successor for the route

The trick is that the metrics compared are calculated in two different nodes:

RA is the local node with two paths to 10.10.10.0/24

RB is a candidate feasible successor that passes the feasibility condition :

RB best metric to 10.10.10.0/24 (=advertised distance) is better then local router RA best metric to 10.10.10.0/24

RB cannot be the feasible successor because RC is a better choice.

Hope to help

Giuseppe

carl_townshend Fri, 04/03/2009 - 07:46

Hi Guiseppe

when advertising the distance from the lan, will it include the lan and wan interface and then send these to router A, or will it just be the BW and DLY from the wan interface ?

also wht are the exact calculations ?

Giuseppe Larosa Fri, 04/03/2009 - 07:56
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Hello Carl,

in my example the advertised distance is only calculated using the LAN interface parameters.

This is really the key point: the advertised distance is the metric as seen by RB and RC and doesn't include the parameters of the WAN interface to RA.

Then RA to calculate the feasible distance needs to take in account the BW and delay of the wan interfaces towards RB and RC.

So for the other thread when I said to increase delay, it has to be increased on RA on interface to RC to make RC less preferred then RB.

the delays on RB and RC are considered for calculating the metric of networks connected to RA.

I tried to provide the real calculations I've just used easy numbers for delay and BW.

Hope to help

Giuseppe